I took a bucket, drilled 2 different sized holes on the side near the bottom and filled it with water. The stream of water the proceeded from the larger hole traveled further than the stream from the smaller one. How does the size of the hole affect the distance that the water travels?
See also: Torricelli's Law.
Answer
I believe the physically relevant parameters here are: 1. internal pressure P at the bottom of the bucket but not too close to the hole, units [$kg*m^{-1} s^{-2}$], the fluid density $\rho$ [$kg*m^{-3}$], hole area $A$ units $m^2$, and viscosity $\eta$ $kg m^{-1} s^{-1}$. First, let's ignore the viscosity, and find which combinations can give us a velocity, units [$m/s$]. The unique combination is $\sqrt(P/\rho)$, and so in the limit of vanishing viscosity the velocity will be independent of the hole size. Now if we add in viscosity, there is a dimensionless combination available, $\rho \eta^2 P^{-1} A^{-1}$. We could have any function of this dimensionless ratio, but common sense tells us that at least in the limit of small $\eta$ it should be a decreasing function, since viscosity should take kinetic energy away from the fluid. From this ratio we see that decreasing the size of the hole is equivalent to increasing the viscosity, and thus the smaller hole should have a smaller exit velocity than the bigger one. I guess this is a long-winded way of saying the big hole will have higher Reynolds number and be less effected by the viscosity.
No comments:
Post a Comment